The area of a rectangle is 45f * t ^ 2 , and the length of the rectangle is 1 ft more than twice the width. Find the dimensions of the rectangle.

[tex]\begin{gathered} \text{ Area of the rectangle = length }\times\text{ width} \\ \text{ Area = 45ft}^2 \\ \text{ 45 = \lparen2w + 1\rparen}\times\text{ w} \\ \text{ open the parenthesis} \\ \text{ 45 = 2w}^2\text{+ w} \\ \text{ 2w}^2\text{ + w - 45 = 0} \end{gathered}[/tex]

Step 4: Solve the quadratic function in step 3 using the general method

[tex]\begin{gathered} \text{ The general form of the quadratic method is written below as} \\ \text{ ax}^2\text{ + bx + c = 0} \\ \text{ Since the quadratic function is 2w}^2\text{ + w - 45} \\ \text{ Relating the two functions, we have} \\ \text{ a= 2} \\ b\text{ = 1} \\ \text{ c =- 45} \\ \text{ } \end{gathered}[/tex][tex]\begin{gathered} \text{ The general formula is } \\ \text{ x= }\frac{-b\text{ }\pm\sqrt{b^2\text{ - 4ac}}}{2a} \\ x\text{ = w} \\ \text{ w = }\frac{-(1)\pm\sqrt{1^2\text{ - 4}\times2\times(-45)}}{2\times2} \\ w\text{ = }\frac{-1\text{ }\pm\sqrt{1^2\text{ -\lparen-360\rparen }}}{4} \\ w\text{ = }\frac{-1\pm\sqrt{1\text{ + 360}}}{4} \\ w\text{ = }\frac{-1\pm\sqrt{361}}{4} \\ w\text{ = }\frac{-1\pm19}{4} \\ \text{ w =}\frac{-1\text{ + 19}}{4}\text{ or }\frac{-1\text{ - 19}}{4} \\ w\text{ = }\frac{18}{4}\text{ or }\frac{-20}{4} \\ \text{ w = 4.5 ft or -5ft} \end{gathered}[/tex]

Hence, the width of the rectangle is 4.5ft

Step 5: Find the length of the width by putting w = 4.5

[tex]\begin{gathered} \text{ Recall, that } \\ \text{ l = 2w + 1} \\ \text{ w= 4.5} \\ \text{ l = 2\lparen4.5\rparen + 1} \\ l\text{ = 9 + 1} \\ \text{ l = 10 ft} \end{gathered}[/tex]

Hence, the length of the rectangle is 10ft